Distribution and Moments of the Brownian Self-Integral

From the Self-Integral to a Normal Variable

Fix a time horizon \(T>0\) and work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). The Brownian self-integral on \([0,T]\) is the Ito integral with integrand \(H_s=B_s\). This integrand is adapted and progressively measurable, and it belongs to the class \(\mathcal{L}^2\), since

Thus \(I_T=\int_0^T B_s\,dB_s\) is well-defined as a square-integrable Ito integral. Its key identity comes from quadratic variation, not the ordinary chain rule. On a partition \(0=t_0<\cdots<t_n=T\), write \(\Delta B_i=B_{t_i}-B_{t_{i-1}}\). The exact algebraic identity \(B_{t_i}^2-B_{t_{i-1}}^2=2B_{t_{i-1}}\Delta B_i+(\Delta B_i)^2\), summed over the partition, gives \(B_T^2=2\sum_iB_{t_{i-1}}\Delta B_i+\sum_i(\Delta B_i)^2\). As the mesh tends to zero, the left-endpoint sums converge to the Ito integral, and the quadratic-variation sums converge to \(T\). Consequently, the identity established from quadratic variation is

Since \(B_T\) has the normal distribution \(N(0,T)\), we can write \(B_T=\sqrt{T}\,Z\) in distribution, where \(Z\sim N(0,1)\). Therefore

This representation determines the entire distribution. It also warns against a common shortcut: a stochastic integral with a random integrand need not be Gaussian, even though deterministic Wiener integrals are Gaussian.

Distribution and Moment-Generating Function

Proof. Since \(Z^2\ge0\), the representation \(I_T\overset{d}{=}(T/2)(Z^2-1)\) implies \(I_T\ge -T/2\). Equality would require \(Z=0\), an event of probability zero for a standard normal variable, so there is no atom at the boundary. For \(y>0\), the transformation \(Y=Z^2\) has density \(f_Y(y)=\dfrac{1}{\sqrt{2\pi y}}e^{-y/2}\): the two preimages are \(z=\sqrt y\) and \(z=-\sqrt y\), and each contributes the normal density multiplied by \(\left|dz/dy\right|=1/(2\sqrt y)\). Now set \(y=1+2x/T\), so \(dy/dx=2/T\). Substitution into \(f_Y(y)\) gives the displayed density for \(x>-T/2\). For the moment-generating function, when \(\lambda<1/T\), the normal integral for \(\mathbb{E}[e^{\lambda T Z^2/2}]\) is finite and equals \((1-\lambda T)^{-1/2}\); multiplying by \(e^{-\lambda T/2}\) gives the formula. If \(\lambda\ge1/T\), the Gaussian integral diverges, so the moment-generating function is not finite there. \(\square\)

The moment-generating function is useful when exponential moments are wanted, but its domain matters: the positive exponential moment exists only for \(\lambda<1/T\). In particular, one should not assume that every positive exponential moment is finite merely because all polynomial moments are finite.

Expectation, Variance, and All Raw Moments

Proof. For a standard normal variable, integration by parts in its density gives \(\mathbb{E}[Z^{2k}]=(2k-1)\mathbb{E}[Z^{2k-2}]\) for \(k\ge1\). The boundary term vanishes because a polynomial times \(e^{-z^2/2}\) tends to zero at both ends of the real line. Starting from \(\mathbb{E}[Z^0]=1\), induction yields \(\mathbb{E}[Z^{2k}]=(2k-1)!!\). Expanding \((Z^2-1)^n\) by the binomial theorem and taking expectations proves the general formula. For \(n=1\), it gives \((T/2)(1-1)=0\). For \(n=2\), it gives \((T/2)^2(3-2+1)=T^2/2\); since the mean is zero, this is the variance. For \(n=3\), the sum is \(15-9+3-1=8\), giving \(T^3\). For \(n=4\), the sum is \(105-60+18-4+1=60\), giving \(15T^4/4\). \(\square\)

The expectation and variance are consistent with the Ito integral itself. In fact, the Ito isometry gives \(\mathbb{E}[I_T^2]=\mathbb{E}\int_0^T B_s^2\,ds=T^2/2\), and the integral has mean zero. More specifically, \(M_t=\int_0^tB_s\,dB_s\) is a true martingale on any fixed finite horizon: it is square-integrable because \(\mathbb{E}[M_t^2]=t^2/2<\infty\), and its conditional expectation satisfies \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) for \(s\le t\). This conditional-mean property follows from the Ito integral construction: for adapted step integrands, each future Brownian increment has conditional mean zero given its left endpoint, and the property passes to the square-integrable Ito-integral limit. The martingale property explains the zero expectation, but does not by itself determine the distribution.

Common Pitfall: The Isometry Does Not Give the Law

A tempting but incorrect calculation is to treat \(\int_0^T B_s\,dB_s\) as Gaussian just because the Ito isometry supplies a variance. The isometry does give \(\mathbb{E}[I_T^2]=T^2/2\), but a mean and variance do not determine a distribution. For example, a mistaken conclusion would be \(I_T\sim N(0,T^2/2)\); that Gaussian law has third moment zero. The actual third moment is \(T^3\), as follows from the shifted-square representation. The correction is to use the quadratic-variation identity to identify the random variable itself, and only then calculate its law and moments.

The integral is centered and is a true martingale at each finite time, yet its terminal distribution is asymmetric and supported below only by the finite boundary \(-T/2\). These facts are compatible: the martingale property constrains conditional means, not Gaussianity or symmetry. The distinction is important whenever the integrand depends on the Brownian path being integrated against.